@utkarsh_2105 It’s even easier, you don’t need to prove halting’s undecidable. The number of unary languages is uncountable; the decidable languages are countable
Important complexity consequence of new claimed quasireimann (https://t.co/OIONdprYfI): Valiant's conjecture VP != VNP over the complex numbers follows from the boolean-complexity lower bound NP not in P/poly. Burgisser proved this under GRH. (🧵1/4)
coefficients, which have solutions over C, have solutions modulo many "reasonably-sized" primes, where "reasonably-sized" has some precise dependence on the # variables, degree, and coefficient magnitude of the system (but no dependence on # of equations) (4/4).
Burgisser showed we can choose some prime p (small enough), replace all the complex constants with some F_p constants, and magically the computation still works (on 0/1 inputs). This follows from a very general statement: polynomial systems of equations with integer ... (3/4)
@profimoto It’s not quite black-box since their o(nlogn) bound isn’t in the circuit model. For FFT it’s more or less black-box. Nonetheless their construction seems to involve a direct counterexample inside of it which is isolated and simplified here
My loyal servant has extracted a quite simple counterexample to the Network Coding Conjecture from OpenAI's new integer multiplication algorithm. Link here: https://t.co/O6Ql24jl75
Afghani et al cast this result as an avenue towards circuit lower bounds, but funnily enough it has ended up yielding a disproof of the network coding conjecture instead. such is life
Previous work by Afghani, Frekson, Kamma and Larsen (2019) showed that this conjecture implies an nlogn circuit lower bound for multiplication, which is why I've spent some time thinking about it.
@ericneyman The linked paper is considering: given a random truth table of a boolean function, can you prove that it requires large circuit complexity? A counting argument shows that high circuit complexity happens with overwhelming probability (somewhat unlike your assumption, where you…
@henryquantum It is equivalent to Nexp in P/poly. The same holds for other similar nonuniform separations, I.e. np/poly vs p/poly. The relevant fact is that NEXP contains a language (e.g. succinct circuit sat) which is complete for NEXP/poly under P/poly reductions.